$\mathbb C/\Lambda$ is isomorphic $\mathbb C/\Lambda'$ if and only if $\Lambda' = c \Lambda$ for some $c \in \mathbb C^*$

Question

Note: By isomorphism/automorphism, I mean isomorphism/automorphism of Riemann surfaces. $\Lambda,\Lambda'$ are discrete lattice in $\mathbb C$.

Backward direction:

$z\mapsto c z$ gives an automorphism of $\mathbb C$ which drops down to an isomorphism $\mathbb C/\Lambda \rightarrow \mathbb C/\Lambda'$.

Forward direction:

This is where I am stuck. So far I have argued:
Let $f:\mathbb C/\Lambda \rightarrow \mathbb C/\Lambda'$ be the given isomorphism and let $\pi: \mathbb C \rightarrow \mathbb C/\Lambda$ and $\pi': \mathbb C \rightarrow \mathbb C/\Lambda'$ be projection maps.
WLOG $f([0])=[0]$, because, otherwise, we can postcompose by a translation.

Let $N_0$ be a small enough nbd of $0$ in $\mathbb C$.
As $f,\pi$ are local isomorphisms, $f\circ \pi: N_0 \rightarrow f(\pi(N_0))$ is an isomorphism.
As $\pi'$ is a covering map, there is a nbd, $N_0'$ of $0$ in $\mathbb C$ such that $(\pi')^{-1}: f(\pi(N_0)) \rightarrow N_0'$ is an isomorphism.

Now we proceed to show $(\pi')^{-1} \circ f\circ \pi : N_0\rightarrow N_0'$ is given by $z\mapsto c z$ for some $c\in \mathbb C\backslash 0$.

Let $\eta$ be a non-vanishing holomorphic 1-form on $\mathbb C/\Lambda'$ such that $(\pi')^*(\eta)=dz$.
$\pi^*(f^*(\eta))$ is a non-vanishing $\Lambda$-periodic and holomorphic 1-form on $\mathbb C$.
By Liouville's theorem, we are forced to have $\pi^*f^*(\eta)=cdz$ for some $c\in \mathbb C\backslash 0$.

Thus, $\pi^*f^*(\pi'^{-1})^*(dz) = cdz$.
This gives $\pi'^{-1} \circ f \circ \pi: N_0 \rightarrow N_0'$ is $z\mapsto cz$ as derivative is a constant $c$ and $0$ maps to $0$.

How do I use the above to show that $f([z])=[cz]$?

Answer 1

It's an explanation(copy) of Function's excellent comment and answer, which is in fact a complete proof. I'm learning Riemann surface too. As an exercise, I am trying to unravel the details.

Mainly there is five components:

1. The lifting criterion and the unique lifting property for "topological" covering space.
2. Moving the lemmas to "analytic" covering spaces.
3. manupulating the commutative diagram to get the lifting $\widetilde{f}: \Bbb{C} \to \Bbb{C}$ and to conclude it's an automorphism.
4. $\widetilde{f}(z) = cz$ for some $c \neq 0$ via How can it be shown that $\mathrm{Aut}(\mathbb{C})=\{f\,|\,f(z)=az+b,a\neq 0\},$ is defined as bijective ...
5. Conclude $\Lambda^\prime = c \Lambda$

The lifting criterion and the unique lifting property, stated in Hatcher's Algebraic Topology, page 62, is:

Proposition 1.33.(The lifting criterion) Suppose given a covering space $p: (\widetilde{X}, \widetilde{x}_0) \to (X, x_0)$ and a map $f: (Y, y_0) \to (X, x_0)$ with $Y$ path-connected and locally path-connected. Then a lift $\widetilde{f}: (Y, y_0) \to (\widetilde{X}, \widetilde{x}_0) $ of $f$ exists iff $f_*(\pi_1(Y, y_0)) \subset p_*(\pi_1(\widetilde{X}, \widetilde{x}_0))$

Proposition 1.34.(The unique lifting property) Given a covering space $p: \widetilde{X} \to X$ and a map $f: Y \to X$, if two lifts $\widetilde{f}_1, \widetilde{f}_2: Y \to \widetilde{X}$ of $f$ agree at one point of $Y$ and $Y$ is connected, then $\widetilde{f}_1$ and $\widetilde{f}_2$ agree on all of $Y$.

But here all the spaces are topological manifolds and all the maps are continuous maps. So we need some more lemmas to move the results to complex manifolds(Here I state them in context of Riemann surfaces, but they holds on complex manifolds or smooth manifolds):

(Lemma 1, analytic covering space) Suppose given a Riemann surface $X$ and a (topological) covering space $p: (\widetilde{X}, \widetilde{x}_0) \to (X, x_0)$, then there is an unique atlas on $\widetilde{X}$ making $\widetilde{X}$ a Riemann surface such that $p$ is holomorphic.

(Lemma 2, the holomorphic lifting) Suppose given an analytic covering space $p: (\widetilde{X}, \widetilde{x}_0) \to (X, x_0)$, and a holomorphic map $f: (Y, y_0) \to (X, x_0)$. If a lift $\widetilde{f}: (Y, y_0) \to (\widetilde{X}, \widetilde{x}_0) $ of $f$ exists and is continous, then it's also holomorphic.

So these two lemmas move results from continous context to holomorphic context. For details of these stuff, please see Mosher's answer in the universal covering space of compact Riemann surface(how the covering map be holormporphic?), or John M. Lee's Introduction to Smooth Manifolds section 4.3, smooth covering maps, page 92. Or the vedio The Riemann Surface Structure on the Topological Covering of a Riemann Surface by nptelhrd on youtube.

Then it's a routine task to analysis the commutative diagram(I'm not quite familiar with writing commutative diagram in latex. Hence I post it as a screenshot.):

Suppose $g$ is the inverse of $f$ and $f([0]) = [0]$. Since $\Bbb{C}$ is simply connected, by the lifting criterion we know there is a unique lifting $\widetilde{f}$ of $f\circ \pi_\Lambda$ such that $\widetilde{f}(0) = 0$, and by lemma 2 "the holomorphic lifting" $\widetilde{f}$ is holomorphic. Similarly there is a unique lifting $\widetilde{g}$ of $g \circ \pi_{\Lambda^\prime}$ such that $\widetilde{g}(0) = 0$, and by lemma 2 "the holomorphic lifting" $\widetilde{g}$ is holomorphic.

Hence $\pi_{\Lambda}\circ \widetilde{g} \circ \widetilde{f} = g\circ f \circ \pi_{\Lambda}$. But we know $g\circ f =1$ since $g$ is the inverse of $f$. So $\pi_{\Lambda}\circ \widetilde{g} \circ \widetilde{f} = \pi_{\Lambda}$. Hence $\widetilde{g} \circ \widetilde{f}$ is a lifting of $\pi_{\Lambda}$ such that $(\widetilde{g} \circ \widetilde{f})(0)=0$. But $1$ is also such an lifting. Hence by the unique lifting property we have $\widetilde{g} \circ \widetilde{f} = 1$. Similarly $\widetilde{f} \circ \widetilde{g}=1$. Hence $\widetilde{f}$ is an automorphism of $\Bbb{C}$.

By 4. and our assumption $\widetilde{f}(z)=cz$ for some $c \neq 0$.

Then:

$$ \begin{aligned} \Lambda^\prime &= \pi_{\Lambda^\prime}^{-1}([0]_{\Bbb{C}/\Lambda^\prime}) \\ &= \widetilde{f}(\widetilde{f}^{-1}(\pi_{\Lambda^\prime}^{-1}([0]_{\Bbb{C}/\Lambda^\prime}))) \\ &= \widetilde{f}(\pi_{\Lambda}^{-1}(f^{-1}([0]_{\Bbb{C}/\Lambda^\prime}))) \\ &= \widetilde{f}(\pi_{\Lambda}^{-1}([0]_{\Bbb{C}/\Lambda})) \\ &= \widetilde{f}(\Lambda) \\ &= c \Lambda \end{aligned} $$

Hence I think this is what Function talking about in the comment and the answer. And I have two remark about this:

1. There should be someway to construct $\widetilde{f}$ directly, that avoids using the lifting criterion, but use something like forms and integral. But I don't know how to do it.
2. the calculation of $Aut(\Bbb{C})$ is kind of non-trivial. At least it's harder than $Aut(\Bbb{D})$. I don't know if there is someway to proof $\Lambda^\prime = c \Lambda$ avoiding this result.

Answer 2

This is supposed to be a comment but is a bit long, so I post it as an "answer" anyway.

The situation is depicted as a commutative diagram \begin{equation*} \begin{matrix} \ \ \mathbb{C} & \dashrightarrow & \mathbb{C}\ \ \\ p\downarrow & &\downarrow q\\ \mathbb{C}/\Lambda & \underset{f}\longrightarrow & \mathbb{C}/\Lambda' \end{matrix} \end{equation*} One starts with the arrow $f$, the vertical arrows are the quotient maps, which are also covering maps. One would like to complete the diagram by a dashed arrow.

This is where one uses the lifting property. Recall the lifting theorem says the following: if one is given the following diagram of "nice" topological spaces (semi-locally simply connected, or "espace délaçable" in Bourbaki's terminology) $$ \begin{matrix} & & E \ \ \ \\ & & \downarrow p\\ S & \underset{f}\rightarrow & B\ \ \ \end{matrix} $$ and if $f_{*}(\pi_1(S)) \subset p_{*}\pi_1(E)$, then there exists a (necessarily unique, if base points are specified) map $\widetilde{f}:S \to E$ such that $p\circ \widetilde{f}=f$. See Munkres, "Topology", Lemma 79.1.

Back to the "torus" situation. The map $f\circ p$ is a map from a simply connected space to $\mathbb{C}/\Lambda'$. Its image is trivially contained in the image of $q_*(\pi_1(\mathbb{C}))$, which is equally trivial. Whence it lifts to a unique map $\widetilde{f}:\mathbb{C} \to \mathbb{C}$, sending $0$ to $0$, such that $q\circ \widetilde{f} = f\circ p$. Now one can use the argument stated in my comment.